Geometry Problem on $\triangle ABC$ and Angle Chasing 
$\triangle ABC$ is an isosceles triangle with $AB=BC$ and $\angle ABD=60^{\circ}$, $\angle DBC=20^{\circ}$ and $\angle DCB=10^{\circ}$. Find $\angle BDA$.

My approach: Let $\angle BDA=x$. Let $AB=BC=p$. Applying sine law in $\triangle ADB$, $\dfrac{p}{\sin x}=\dfrac{BD}{\sin (60+x)}$. Applying sine law in $\triangle BDC$, $\dfrac{p}{\sin150^{\circ}}=\dfrac{BD}{\sin 10^{\circ}}$. Using the two equations, we get $\dfrac{1}{2\sin 10^\circ}=\dfrac{\sin x}{\sin (60^\circ +x)} \implies 2\sin 10^\circ=\dfrac{\sqrt{3}}{2}\cot x + \dfrac{1}{2} \\ \implies x = \text{arccot} \left(\dfrac{4\sin 10^\circ-1}{\sqrt{3}}\right)$. 

Now I am stuck. I know that the answer is $100^\circ$ but no matter how hard I try I cannot seem to simplify it any further. Please help. If anybody has a better solution (involving simple Euclidean Geometry), I would be grateful if you provide it too.
Edit: I am extremely sorry. The original problem was when $AB=BC$. Sorry for the inconvenience caused. I have rectified my mistake. Also, I have changed the answer to $100 ^\circ$.
 A: 
$\angle ABC=\angle ABD+\angle DBC=80^\circ$.
\begin{align*}
AB&=BC\\
\implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ.
\end{align*}
Erect an equilateral triangle $ACE$ on base $AC$.
Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle AEB=\angle BEC=30^\circ.$$
$$\angle CDB=180^\circ-\angle DBC-\angle BCD=150^\circ.$$ Thus quadrilateral $BDCE$ is cyclic because its angles $D$ and $E$ are supplementary. Thus 
$$\angle DEC=\angle DBC=20^\circ.$$
\begin{align*}
\angle ECB&=\angle ECA-\angle BCA=10^\circ\\
\implies \angle ECD&=\angle ECB+\angle BCD=20^\circ=\angle DEC.
\end{align*}
Thus triangle $CED$ is isosceles on base $CE$, so $CD=DE$. Thus $\triangle$s $ACD, AED$ are congruent in opposite sense because $AC=AE$, $CD=ED$ and $AD$ is common. Thus
\begin{align*}
\angle CAD&=\angle DAE=30^\circ\\
\angle BAE&=\angle CAE-\angle CAB=10^\circ\\
\implies \angle DAB&=\angle DAE-\angle BAE=20^\circ\\
\implies \angle BDA&=180^\circ-\angle DAB-\angle ABD=100^\circ.
\end{align*}
A: 
Let $E$ be the circumcenter of $BCD$. Then $\angle BED=2\angle BCD=20^\circ$ and $\angle DEC =2\angle DBC =40^\circ$. Hence $\angle BEC=60^\circ$. This and $BE=EC$ shows that $BEC$ is equilateral. So $BC=BE$ and $\angle CBE=60^\circ$. By assumption $AB=BC$, so $AB=BE$ and $$\angle BEA = 90^\circ -\frac 12 \angle ABE =90^\circ -\frac 12 \cdot 140^\circ =20^\circ =\angle BED.$$ Therefore $A,D,E$ are collinear and we find $$\angle BDA =180^\circ -\angle EDB = \angle BED+\angle DBE= 20^\circ+80^\circ =100^\circ.$$
A: Continue to simplify
$$\begin{align}
\cot x & =\frac{4\sin 10-1}{\sqrt{3}} 
=\frac{(2\sin 10-\frac12)\cos10}{\frac{\sqrt{3}}2\cos10} \\
& =\frac{\sin 20-\cos60\cos10}{\cos10\sin60} 
=\frac{2\cos 70-2\cos60\cos10}{\cot10\cdot2\sin10\sin60} \\
& =\frac{\cos70-\cos50}{\cot10\cdot(\cos50-\cos70)} =-\cot80=\cot100
\end{align}$$
Thus, $x=100^\circ$.
A: Assuming $AB=BC$ is what you intended, your calculation is correct. Notice that $\frac{4 \sin 10^\circ - 1}{\sqrt 3}$ is negative, and in fact the arccot of this value is $-80^\circ$. How can the angle be negative?! Recall that $x$ has to be an obtuse angle, so you should add $180^\circ$ to $-80^\circ$, obtaining $100^\circ$. You can confirm that $x=100^\circ$ also satisfies the equation you obtained.
A: It you are looking for a "clever" way to solve the obtained trigonometric equation, the following trick is often useful in similar problems:
Let $x $ satisfy the equation:
$$
\frac {\sin (x)}{\sin (C-x)}=\frac {\sin (A)}{\sin (C-A)},\quad 0<x,A <C <\pi.\tag1
$$
Then
$$ x=A.\tag2$$
Applying this to your problem one obtains:
$$\frac {\sin (x)}{\sin (120^\circ-x)}=\frac1{2\sin 10^\circ}
=\frac{\cos 10^\circ}{\sin 20^\circ}=\frac{\sin 100^\circ}{\sin 20^\circ}\implies x=100^\circ.
$$

Proof of $(1)\implies (2) $:
$$\begin{align}
&\frac {\sin x}{\sin (C-x)}=\frac {\sin A}{\sin (C-A)}\\
&\iff
\sin x\,(\sin C \cos A-\cos C\sin A)=\sin A\,(\sin C \cos x-\cos C\sin x)\\
&\iff
\sin C\,(\sin x\cos A-\cos x \sin A)=0\\
&\iff\sin C\sin(x-A)=0\stackrel{0<x,A <C <\pi}\implies x=A.
\end{align}
$$
A: Although not as satisfying as a purely geometrical solution, the most direct method is to apply the Trigonometric Form of Ceva's Theorem : 
$$\frac{\sin\alpha}{\sin(A-\alpha)}.\frac{\sin\beta}{\sin(B-\beta)}.\frac{\sin\gamma}{\sin(C-\gamma)}=1$$
where $A, B, C$ are the angles of the triangle which are split by the concurrent cevians into angles $\alpha, A-\alpha, \beta, B-\beta, \gamma, C-\gamma$ in order round the triangle.
The resulting equation of the form $$R\sin\alpha=\sin(A-\alpha)$$ has the solution $$\tan\alpha=\frac{\sin A}{R+\cos A}$$
In your problem 
$$R=\frac{\sin40^{\circ}}{\sin10^{\circ}}.\frac{\sin20^{\circ}}{\sin60^{\circ}}=1.4619022$$ $$\tan\alpha=\frac{\sin50^{\circ}}{1.4619022+\cos50^{\circ}}=0.36397$$ $$\alpha=20^{\circ}$$
$$\angle BDA = 180^{\circ}-60^{\circ}-\alpha=100^{\circ}$$
